Saturday, December 30, 2006

example usage of Visual Thesaurus, and Thinkmap in general


south-west to north-east. It represents from broad to narrow:

the humanities -> philosophy -> branches (& literally!) of philosphy

the underlying engine powering Visual Thesaurus is Thinkmap
you can also try Visual Thesaurus online at http://visualthesaurus.com

have fun!

Visual Thesaurus

YOUR THESAURI ARE NOT ENOUGH!!!

SERIOUSLY check this out...see that in the box?
some words seem smaller and faded because the words are actually surrounding the center like a ball
it is in 3D

words are link to other words as thoughts are linked to other thoughts
isolated information don't live too long in the accessible memory of our brains
therefore, the best way to learn about words is to learn them and their meanings, in association with other words and meanings
Visual Thesaurus lays it out for you, the relations between the words and their corresponding senses
relations include:
  • antonym
  • is a particle of
  • is derived from
  • pertains to
  • see also
  • is similar to
  • attribute
  • entails
  • is a type of
  • is a part of
  • is made of
  • is a member of
  • verb group
  • domain category
  • domain region
  • domain usage
i think it uses the Princeton Wordnet database, which is great
having it mapped out like this in Visual Thesaurus is just fucking cool, not to mention useful. (the other way around)

so what are you waiting for?
get it here (note: illegal copy)
http://rapidshare.de/files/9901279/Thinkmap.Visual.Thesaurus.v3.01.Build.1222.Desktop.Edition.Regged.rar
password:avax

and then thank yourself

Wednesday, December 13, 2006

how to temporarily offset myopia

for those of you who are short-sighted, this is just for fun, as it is for those who are not as well...

step 1 for all methods - take off your glasses!

  • method 1. - look through a very tiny hole at close range, right next to your eye. while this limits your field of view, it makes it much clearer and in focus.
  • method 2. - look at a convex reflective object at close range, all though this will make them small, they will be in focus
  • method 3. - on your eyelid covering the top part of your eye(s), gently press with your fingers onto your eyeball(s) (you'll have to apply quite a bit of pressure), while keeping your eye(s) open. this puts the retina at the point of focus by compressing your eyeball(s). (myopia is due to accommodation by the lengthening of the eyeball.

hopefully you weren't careless enough to damage your eye(s). and tasted a moment's glimpse, however short, of non-short sightedness...

of course, if your eyes weren't myopic, then you would have had a glimpse of short-sightedness. now go apologize to that kid you bullied back in primary school...

i would be really interested to hear of any more methods not already listed here. please let us all share these pathetic fantasies of 20/20 vision.

Thursday, December 07, 2006

a solution to Bertrand's Paradox?

I have formulated a tentative possible solution. I leave for the mathematicians to judge.
As my math really sucks, what I did was what I tend to be better at, which is intuitive logic.
So here is a logical attempt at a solution to Bertrand's paradox.

Please refer to the Wikipedia article, I use their numbering of the methods of solution.

Fig. A

Fig. A shows method 1

  • It is only necessary to consider the left half of the circle.
  • The midpoints of the chords AC, AB and AD, which are P, Q and R, lie on the circumference of the inner smaller circle.
  • The midpoints of possible chords which are longer than a side of the inscribed equilateral triangle lie on the arc QO, which is 1/3 of the left half of the circumference of the inner circle.
  • The midpoints of possible chords which are shorter, lie on an arc AQ which is 2/3 of the left half of the inner circle.

It would seem obvious that the probability of the length of a chord randomly chosen, to be longer than a side of the triangle is 1/3. This solution however, is incompatible with methods 2 & 3.

Here is a possible reconciliation.
  • The problem is determining the probability of the chord being longer.
  • A chord has 2 properties. Length & Orientation.
  • Therefore, there cannot possibly be chords without either or neither. i.e. chords that have orientation but no length, chords that have length but no orientation, or chords that have neither, cannot exist. viz. All chords have both Length and Orientation
  • For a chord with x length, there are infinite possible orientations.
  • For a chord with orientation y, there are infinite possible lengths from zero to the length of the diameter.
  • Orientation can be held constant, i.e. for each length there is equal infinite possible orientations.
  • We can do this by fixing orientation, i.e. consider only one orientation.
  • This is in fact what is done in method 2.
  • All the midpoints of chords considered in method 2 lies along the radius which is a straight line. Therefore all of their orientations are the same.

In Fig. A the the midpoints of chords with different length and orientation form the arc a2 and b2. Because the chord is projected onto the outer circle, and not a circle with a center point at A, this distorts the correspondence between length and orientation. While orientation changes constantly with each degree of angle, the lengths corresponding to each successive orientation does not. Each degree thus represents different proportions of possible lengths. The result being disproportionately more chords (shorter) are represented by the arc b2, than by a2 (longer). But orientation was held constant. This calculation results in a distorted false probability.

A correction can be applied to the resulting false probability of 1/3 to yield the correct (i think) probability of 1/2, which can be found by method 2.

Note: In the following calculation, D is the diameter




Method 3 uses the relative distribution of midpoints in and outside of the inner circle to acquire a probability of 1/4.

Fig. B
Fig. B shows method 3.
  • The areas of the 2 circles, can be represented by 2 rectangles of similar area .(see animation)
  • The formula for the area of a circle is pi*r^2.
  • The length of the rectangle is pi*r.
  • The height of the rectangle is r.
  • Therefore its area is equal to the larger circle.
  • The reasoning is similar for the smaller circle.
  • The height (r) can also be thought of as a dimension, which represents length of chord (by virtue of correspondence to a visualization of Archimedes' proof).
  • The length (pi*r) can similarly represent orientation.
  • Any point within the larger rectangle has coordinates representing length and orientation.
  • The smaller circle, represented by the smaller rectangle, claims to contain midpoints of chords longer, and therefore all possible chords longer, than a side of the triangle.
  • The smaller rectangle proves otherwise. The height of the smaller rectangle being half of the length of the larger rectangle represents correctly half of the possible lengths of the chord being longer than a side of the triangle. However the area of the smaller rectangle does not take into account all possible orientations of those chords. Namely only half. As the length of the smaller rectangle is half of the big one. This unfortunately results in a probability of 1/4, which is half of that of a correct solution.
click on image to view animation: visualization of Archimedes' proof (by Bettina Richmond)

representation of areas of circles in method 3. (not drawn to scale)


But why does calculating the area of the smaller circle leave out half of the original orientations taken into account by the area of the larger circle?

The next diagram makes it obvious.
The arc ABC is twice as long as the corresponding arc of the smaller circle, DEF. However, the orientations they represent are supposed to be the same. The angle SOR and TOU are the same. But the arc SR and TU are different in length. The same goes for the areas a and b.

The arcs ABC and DEF are respectively represented as pi*r and 1/2 pi*r in the rectangles above. We cannot work directly with infinity of course. We don't can't enumerate the magnitude of infinity. We can't count the infinite possible orientations. What we can do however, is keeping it constant as we work with another variable. In this case the length of chord. Methods 1 & 3 have been shown not to have kept orientation constant, while method 2 has. That is why method 2 gives a correct answer.

However, we can also show how off the mark our calculations were. pi*r represents the infinite possible orientations. 1/2 pi*r represents half of that infinity. That is why the probability obtained using method 3 (1/4) is half of what is correct (1/2).


That's it! The probability of the length of a randomly drawn chord in the circle being longer than a side of the inscribed equilateral triangle is not all 1/3, 1/2 and 1/4.

It is 1/2.

And I believe I have proven it, if not mathematically, then logically.

Please do criticize me, as I have said, my math sucks! I might have made a terrible mistake here.

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Friday, December 01, 2006

Jack of all trades, master of none...

i have always hated this phrase.
but now, i realize it is because it is incomplete...

the complete and authentic epithet is:

"Jack of all trades, master of none, though ofttimes better than master of one"

now that is MUCH better. and how true.